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- 2072 Magh · 8 marks
Rigid container two-phase heated from 100 to 150°C
Short answer
At 100°C, pressure 101.4 kPa, vapor mass 0.298 kg. At 150°C, pressure 476.1 kPa, vapor mass 1.273 kg. Heating to saturated vapor at constant volume occurs at 156.4°C.
A closed rigid container of volume 0.5 m³ is placed on a hot plate. Initially, the container holds two phase mixture of saturated liquid water and saturated water vapor at T1 = 100°C with a quality of 0.2. After heating, the temperature in the container is T2 = 150°C. Indicate the initial and final states on P-v and T-v diagrams and determine: a) The pressure at each state b) The mass of the vapor present at each state, in kg c) If the heating continued, determine the temperature, when the container holds only saturated vapor. [Refer the attached tables for properties of steam]
Given: V = 0.5 m³ (rigid, constant v), state 1: T₁ = 100 °C, x₁ = 0.2; state 2: T₂ = 150 °C.
Steam table values:
| T | P_sat (kPa) | v_f | v_g (m³/kg) |
|---|---|---|---|
| 100 °C | 101.4 | 0.001043 | 1.6719 |
| 150 °C | 476.1 | 0.001091 | 0.3925 |
a) Pressures
P₁ = P_sat(100 °C) = 101.4 kPa (≈ 0.101 MPa)
State 2: v is constant. First find v₁:
v₁ = v_f + x v_fg = 0.001043 + 0.2 × (1.6719 − 0.001043) = 0.3352 m³/kg
At 150 °C, v_f < 0.3352 < v_g = 0.3925, so state 2 is still a wet mixture: P₂ = P_sat(150 °C) = 476.1 kPa
b) Mass of vapour at each state
Total mass m = V/v₁ = 0.5/0.3352 = 1.4916 kg
State 1: m_g1 = x₁ m = 0.2 × 1.4916 = 0.298 kg
State 2: x₂ = (v − v_f)/(v_g − v_f) = (0.3352 − 0.001091)/(0.3925 − 0.001091) = 0.8536
m_g2 = 0.8536 × 1.4916 = 1.273 kg
c) Temperature when only saturated vapour remains
Set v_g(T) = 0.3352 m³/kg. From the saturated tables: v_g = 0.3465 at 155 °C and 0.3068 at 160 °C. Interpolating:
T = 155 + 5 × (0.3465 − 0.3352)/(0.3465 − 0.3068) ≈ 156.4 °C (P_sat ≈ 565 kPa)
P|
| 3 (sat. vapour, ≈ 565 kPa)
| 2 (476.1 kPa)
| |
| | all on v = 0.3352 = const
| 1 (101.4 kPa)
+----------------- v
T|
| 3 (≈ 156.4 C, on sat. vapour line)
| 2 (150 C)
| |
| 1 (100 C)
+----------------- v
All three states lie on one vertical line (v = 0.3352). States 1 and 2 are inside the dome, and state 3 is on the saturated vapour line.
✓ Answer checked by Claude Opus 5.5 (high effort)
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